# If Nc4 , Nc5 and Nc6 Are in A.P., Then Find N. - Mathematics

If nC4 , nC5 and nC6 are in A.P., then find n.

#### Solution

Since nC4 , nC5 and nC6 are in AP.

∴ 2. nC5 = nC4 + nC6

$\Rightarrow 2 \times \frac{n!}{5!\left( n - 5 \right)!} = \frac{n!}{4!\left( n - 4 \right)!} + \frac{n!}{6!\left( n - 6 \right)!}$
$\Rightarrow \frac{2}{5 \times 4!\left( n - 5 \right)\left( n - 6 \right)!} = \frac{1}{4!\left( n - 5 \right)\left( n - 4 \right)\left( n - 6 \right)!} + \frac{1}{6 \times 5 \times 4!\left( n - 6 \right)!}$
$\Rightarrow \frac{2}{5\left( n - 5 \right)} = \frac{1}{\left( n - 5 \right)\left( n - 4 \right)} + \frac{1}{30}$
$\Rightarrow \frac{2}{5\left( n - 5 \right)} - \frac{1}{\left( n - 5 \right)\left( n - 4 \right)} = \frac{1}{30}$
$\Rightarrow \frac{2n - 8 - 5}{5\left( n - 5 \right)\left( n - 4 \right)} = \frac{1}{30}$
$\Rightarrow \frac{2n - 13}{\left( n - 5 \right)\left( n - 4 \right)} = \frac{1}{6}$
$\Rightarrow 12n - 78 = n^2 - 9n + 20$
$\Rightarrow n^2 - 21n + 98 = 0$
$\Rightarrow \left( n - 7 \right)\left( n - 14 \right) = 0$
$\therefore n = 7 \text{and} 14$
Concept: Combination
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 17 Combinations
Exercise 17.1 | Q 12 | Page 8